Limits of Accuracy
No measurement is exact. Limits of accuracy describe how far the true value could be from the reading, which depends on the finest marking on the instrument's scale. This NSW Year 9 Core maths subtopic is Stage 5 Core content under NESA outcome MA5‑MAG‑C‑01.
You will read the size of one unit off rulers, thermometers, jugs, scales and speedometers, halve it to get the limit of accuracy, and write the range the true value lies in, so a plank measured as \(85\) cm to the nearest centimetre is between \(84.5\) cm and \(85.5\) cm. You will also work with readings to the nearest tenth or hundred, find percentage errors, explain why a longer measurement with the same ruler is more accurate, and find the largest possible perimeter of a rectangle. It is part of Year 9 Core Surface Area and Volume.
Every question with a fully worked solution.
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Theory
No measurement is exact. The limit of accuracy is half of one unit on the scale, either way. This Year 9 maths page shows how to read one unit off a scale, find the range the true value lies in, work out percentage error and use upper and lower limits in a calculation.
Every measuring instrument has a smallest marked division. This is the size of one unit on the scale, also called the precision. A ruler in millimetres has a unit of \(1\) mm.
The limit of accuracy, or absolute error, is half of one unit. A reading can be out by up to that much either way, so we write it with \(\pm\). For a millimetre ruler it is \(\pm 0.5\) mm.
The true value lies between the lower limit (reading minus the error) and the upper limit (reading plus the error).
The way a number is written also tells you the unit. "To the nearest centimetre" means a unit of \(1\) cm. "Correct to one decimal place" means a unit of \(0.1\).
Percentage error compares the error with the size of the measurement.
| Measurement | One unit | Limit of accuracy | True value lies |
|---|---|---|---|
| \(9\) cm, nearest cm | \(1\) cm | \(\pm 0.5\) cm | \(8.5\) cm to \(9.5\) cm |
| \(3.2\) kg, \(1\) d.p. | \(0.1\) kg | \(\pm 0.05\) kg | \(3.15\) kg to \(3.25\) kg |
| \(70\) g, nearest \(10\) g | \(10\) g | \(\pm 5\) g | \(65\) g to \(75\) g |
How to find the limits of accuracy
- Find one unit: the smallest division on the scale, or the place the value was rounded to.
- Halve it. That is the limit of accuracy, written with \(\pm\).
- Subtract and add it to the reading to get the lower and upper limits.
- For percentage error, divide the absolute error by the measurement and multiply by \(100\%\).
- For a calculation, use all the lower limits together and all the upper limits together.
(i) the size of one unit on the scale
(ii) the limit of accuracy.
(i) The labels go up by \(10\) kg, and there are \(5\) gaps between them.
| \(\text{one unit}\) | \(=\) | \(10 \div 5\) |
| \(\) | \(=\) | \(\mathbf{2} \text{ kg}\) |
(ii) The limit of accuracy is half of one unit, either way.
| \(\text{limit of accuracy}\) | \(=\) | \(\pm \dfrac{1}{2} \times 2\) |
| \(\) | \(=\) | \(\mathbf{\pm 1} \text{ kg}\) |
Count the gaps, not the tick marks.
(i) A rod measured as \(8\) cm, to the nearest centimetre.
(ii) A mass recorded as \(4.6\) kg, correct to one decimal place.
(i) "To the nearest centimetre" means one unit is \(1\) cm, so the limit is \(\pm 0.5\) cm.
| \(8 - 0.5\) | \(=\) | \(7.5\) |
| \(8 + 0.5\) | \(=\) | \(8.5\) |
The true length is between \(\mathbf{7.5}\) cm and \(\mathbf{8.5}\) cm.
(ii) One decimal place means one unit is \(0.1\) kg, so the limit is \(\pm 0.05\) kg.
| \(4.6 - 0.05\) | \(=\) | \(4.55\) |
| \(4.6 + 0.05\) | \(=\) | \(4.65\) |
The true mass is between \(\mathbf{4.55}\) kg and \(\mathbf{4.65}\) kg.
(i) A length of \(25\) cm, to the nearest centimetre.
(ii) A length of \(3.6\) m, correct to one decimal place (answer to two decimal places).
(i) The absolute error is \(0.5\) cm. Divide by the measurement.
| \(\text{percentage error}\) | \(=\) | \(\dfrac{0.5}{25} \times 100\%\) |
| \(\) | \(=\) | \(\mathbf{2\%}\) |
(ii) One unit is \(0.1\) m, so the absolute error is \(0.05\) m.
| \(\text{percentage error}\) | \(=\) | \(\dfrac{0.05}{3.6} \times 100\%\) |
| \(\) | \(=\) | \(1.3888\ldots\%\) |
| \(\) | \(\approx\) | \(\mathbf{1.39\%}\) |
The same \(\pm 0.05\) m error on a \(20\) m length gives only \(0.25\%\). Bigger measurements have smaller percentage errors.
Each side could be up to \(0.5\) m less or more than it was recorded.
| \(\text{length}\) | \(:\) | \(6.5 \text{ to } 7.5 \text{ m}\) |
| \(\text{width}\) | \(:\) | \(2.5 \text{ to } 3.5 \text{ m}\) |
Use the lower limits together, then the upper limits together.
| \(\text{smallest}\) | \(=\) | \(2 \times (6.5 + 2.5) = \mathbf{18} \text{ m}\) |
| \(\text{largest}\) | \(=\) | \(2 \times (7.5 + 3.5) = \mathbf{22} \text{ m}\) |
The recorded perimeter is \(20\) m, but it could really be anywhere from \(18\) m up to \(22\) m.
Common pitfalls
Frequently asked questions
What is the limit of accuracy?
The limit of accuracy is half of the smallest unit on the scale, either way. A ruler marked in millimetres has a limit of accuracy of \(\pm 0.5\) mm.
How do you find the upper and lower limits of a measurement?
Halve the unit the measurement was made to. Subtract it for the lower limit and add it for the upper limit: \(8\) cm to the nearest cm lies between \(7.5\) cm and \(8.5\) cm.
How do you calculate percentage error?
Divide the absolute error by the measurement and multiply by \(100\%\). For \(25\) cm to the nearest cm: \(\dfrac{0.5}{25} \times 100\% = 2\%\).
What is the difference between absolute error and percentage error?
Absolute error is the most a reading can be out by, in the units of the measurement. Percentage error compares that with the size of the measurement, so it shows how accurate it really is.
Why is a bigger measurement more accurate with the same ruler?
The absolute error is the same for both, but dividing it by a bigger measurement gives a smaller percentage error.